Solution
= Solution
Write $x=11+100t$. Reduction of the other <integer congruence> modulo $79$ gives $21t\equiv-3\pmod{79}$. The <Bézout's identity> $19\cdot79-15\cdot100=1$ gives $-15\cdot21\equiv1\pmod{79}$, so $t\equiv45\pmod{79}$. Consequently,
$$
\boxed{x=4511+7900k,\qquad k\in\mathbb Z.}
$$
Indeed, $4511=8+79\cdot57=11+100\cdot45$. The <Chinese remainder theorem> guarantees one class modulo $79\cdot100$, since the two moduli are <coprime>.